Let $R$ be a ring with unity and $M$ a right $R$-module. Recall that $M$ is called a finitely cogenerated module if for every collection $\lbrace A_i \rbrace_{i\in I}$ of submodules with $\bigcap_{i\in I}A_i=0$ there exists a finite subset $J\subset I$ with $\bigcap_{j\in J}A_j=0$.
I need to prove the following result.
If the intersection of every chain of nonzero submodules of $M$ is nonzero, then $M$ is finitely cogenerated.
I made the following proof which is not complete:
Proof. Let $\lbrace B_i \rbrace_{i\in I}$ be a collection of submodules of $M$ with $\bigcap_{i\in I}B_i=0$. Suppose that for every finite subset $J\subset I$, $\bigcap_{j\in J}B_j \neq 0$. If $K:=\lbrace i_1,i_2,\ldots \rbrace \subseteq$, then $B_{i_1} \cap B_{i_2} \supseteq B_{i_1} \cap B_{i_2} \cap B_{i_3} \supseteq \ldots$ is a descending chain of nonzero submodules. By the hypothesis, $\bigcap_{i\in K}B_i \neq 0$. This shows that every countable intersection of nonzero submodules is nonzero.
How can I prove that the intersection $\bigcap_{i\in I}B_i \neq 0$ to get a contradiction.
I need any help. Thanks in advance.
Just continue by induction on the cardinality of the index set $I$. That is, prove by induction on $|I|$ that if $(B_i)_{i\in I}$ is a family of submodules of $M$ whose finite intersections are all nonzero, then $\bigcap_{i\in I} B_i$ is nonzero. You have already shown this when $I$ is countable. In general, suppose that this is known for all cardinalities smaller than $|I|$. You can then write $I$ as the union of a chain of subsets $J_\alpha$, each of which has cardinality less than $I$ (for instance, pick a well-ordering of $I$ of length $|I|$ and let the $J_\alpha$ be the initial segments). Then $\bigcap_{i\in J_\alpha}B_i$ is nonzero for each $\alpha$ by the induction hypothesis, and these modules then form a chain whose intersection is $\bigcap_{i\in I}B_i$.